Question

Giải phương trình:

a) \(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)

b) \(\dfrac{x-291}{1700}+\dfrac{x-293}{1698}+\dfrac{x-295}{1696}+\dfrac{x-297}{1694}=4\)

Giúp mình với!!! Mình cần gấp!!!

Answer 1

https://giaibaitapvenha.blogspot.com/2017/12/toan-lop-8-ai-so-giai-phuong-trinh.html

Answer 2

Nếu lời giải giúp ích được cho bạn, truy cập

https://giaibaitapvenha.blogspot.com/2017/12/en-voi-do-homework-for-you-e-trai.html

Answer 3

a,\(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)

=>\(\dfrac{x-3}{113}+1+\dfrac{x-5}{115}+1=\dfrac{x-7}{117}+1+\dfrac{x-9}{119}+1\)

=>\(\dfrac{x+110}{113}+\dfrac{x+110}{115}=\dfrac{x+110}{117}+\dfrac{x+110}{119}\)

=>(x+110)(\(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\))=0

vì 1/113+1/115-1/117-1/119 khác 0

=>x+110=0=>x=-110

\(\dfrac{x-291}{1700}+\dfrac{x-293}{1698}+\dfrac{x-295}{1696}+\dfrac{x-297}{1694}=4\)

=>\(\dfrac{x-291}{1700}-1+\dfrac{x-293}{1698}-1+\dfrac{x-295}{1696}-1+\dfrac{x-297}{1694}-1+4=4\)=>\(\dfrac{x-1991}{1700}+\dfrac{x-1991}{1698}+\dfrac{x-1991}{1696}+\dfrac{x-1991}{1694}=0\)

=>(x-1991)(1/1700+1/1698+1/1696+1/1694)=0

vì 1/1700+1/1698+1/1696+1/1694 khác 0

=>x-1991=0=>x=1991